2019 Hangdian multi school first 6581 vacation [thinking]

2019 Hangzhou Electric Power Co., Ltd The first scene 6581-Vacation【 thinking 】

subject ：http://acm.hdu.edu.cn/showproblem.php?pid=6581

Reference material ：

https://www.cnblogs.com/wawcac-blog/p/11229277.html
https://www.cnblogs.com/Satan666/p/11273615.html

General meaning ：

share n+1 Vehicles ,0 Car number one is our car , other [1,n] Car No. 1 is the one in front of us , All in one lane , The lane width can only accommodate one car , There is a line on the far right , When the front of our vehicle reaches this line, it is counted as passing , Ask us how long it will take us to get to this line at least . The title will give the starting time of each car , The distance from this line s , Maximum speed v, The length of the body l , Each car can drive close to the bottom of the car in front at most , Not more than .

Ideas ：

Read a sentence from someone else's blog, which is very good ：

Squeeze all the cars together , Become a train with different carriages , Then we drive our own car , Push the train forward , Until our car reached the traffic lights ( The goal is ), Then calculate the time it takes for each car to reach its current position at its highest speed , Take the longest one in all cars （ It will get stuck in the back ）.

Code ：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1e5+5;
int l[maxn], s[maxn], v[maxn];
double t[maxn];

int main() {
    
	int n;
	ll sum = 0;
	while(~scanf("%d",&n)) {
    
		for(int i = 0; i < n+1; i++) {
    
			scanf("%d",&l[i]);
			sum += l[i];
		}
		sum -= l[0];
		for(int i = 0; i < n+1; i++)
			scanf("%d",&s[i]);
		for(int i = 0; i < n+1; i++)
			scanf("%d",&v[i]);
		
		double mymax = (s[n] + sum) *1.0 / v[n]*1.0;
		for(int i = n; i >= 0; i--) {
    
			int temp = s[i] + sum;
			t[i] = temp*1.0 / v[i];
			sum -= l[i];
			if(t[i] > mymax) {
    
				mymax = t[i];
			}
		}
		printf("%.10lf\n",mymax);
		mymax = 0;
		sum = 0;
	}
	return 0;
}